6 The Sylow Theorems

The Sylow Theorems are named after the Norwegian mathematician Ludwig Sylow, who published them in 1872. They are the most important structure theorems of finite group theory after Lagrange’s Theorem. Several distinct lines of proof are now known; we shall adopt a blend of the traditional approach, which uses the class equation, and the more modern approach using group actions.

6.1 Conjugates, Centralisers and the Class Equation

We begin with a definition:

Definition 6.1 (Conjugate) Let \(G\) be a group and let \(x, g \in G\). The conjugate of \(x\) by \(g\) is the element \(gxg^{-1}\).

Example 6.1

Consider the group \(D_3\) . The conjugates of \(s_1\) are \[\begin{align*} es_1e^{-1} &=s_1; \ r_1s_1r_1^{-1} = s_2;\ r_2 s_1r_2^{-1}=s_3 \\ s_1s_1s_1^{-1} &=s_1; \ s_2s_1s_2^{-1} = s_3;\ s_3 s_1s_3^{-1}=s_2. \end{align*}\] The set of conjugates of \(s_1\) is therefore \(\{s_1,s_2,s_3\}\). These are also the conjugates of \(s_2\) and \(s_3\).

Clearly \(e\) is its only conjugate since \(ge\ginv = e\) for all \(g \in D_3\). For \(r_1\) we have: \[\begin{align*} er_1e^{-1} &=r_1; \ r_1r_1r_1^{-1} = r_1;\ r_2 r_1r_2^{-1}=r_1 \\ s_1r_1s_1^{-1} &=r_2; \ s_2r_1s_2^{-1} = r_2;\ s_3 r_1s_3^{-1}=r_2. \end{align*}\] The conjugates of \(r_1\) are therefore \(r_1\) and \(r_2\) — these are also the conjugates of \(r_2\).

At this point the traditional approach would be to show that conjugacy defines an equivalence relation on the elements of \(G\) and that the size of each equivalence class divides the order of \(G\). All of this will follow from the Orbit-Stabilizer Theorem once we have established the following:

Lemma 6.1 Let \(\go\) be a group and define \(*\) on the elements of \(G\) by \[\forall \, x, g \in G, \; g*x = gxg^{-1}.\] Then \(*\) is a group action of \(\go\) on \(G\).

Proof.

Clearly \(e \ast x = exe^{-1} = x\) for all \(x \in G\). Moreover, for \(g_1, g_2 \in G\) we have: \[\begin{align*} (g_1g_2)\ast x &= g_1g_2 x g_2^{-1}g_1^{-1} = g_1(g_2 x g_2^{-1})g_1^{-1}\\ &= g_1(g_1 \ast x) g_1^{-1} = g_1 \ast(g_2 \ast x). \end{align*}\] This complete the verifications. We conclude that \(\ast\) is a group action.

If we have established a group action then we know that we must have orbits and stabilisers. The orbits of the above group action are called the conjugacy classes of \(G\), and we denote the conjugacy class containing \(x\) by \(conj_G (x)\), that is \[conj_G(x) = \{ gxg^{-1} \, | \, g \in G \}.\]

Example 6.2

Revisiting Example 6.1 with \(G= D_3\), we have: \[\begin{align*} \conjg(e) &= \{e\} \\ \conjg(r_1) &= \conjg(r_2) = \{r_1,r_2\} \\ \conjg(s_1) &= \conjg(s_2) = \conjg(s_3) = \{s_1,s_2,s_3\}. \end{align*}\] Notice that \[\{\{e\},\{r_1,r_2\}, \{s_1,s_2,s_3\} \}\] is a partition of \(G\).

Although the size of the conjugacy classes differ, they all divide the order of \(G\).

The Orbit-Stabilizer Theorem tells us that the order of each conjugacy class divides the order of the group. We also observe that an element \(x\) belongs to a conjugacy class of size 1 if and only if \(gxg^{-1} = x, \; \forall \, g \in G\) (or, equivalently, if and only if \(x\) is in the centre of \(G\), as \(xg=gx \ifif x = gxg^{-1}\)). As the conjugacy classes partition \(G\) it is a trivial observation to note that the sum of the sizes of the distinct conjugacy classes of \(G\) equals the order of \(G\). This and the preceding remarks form the basis of the so-called Class Equation of \(G\). First we state those observations formally. If \(C_1, C_2, \ldots, C_r\) are the distinct conjugacy classes of a finite group \(G\), then it follows that \[|G| = \sum_{i=1}^r |C_i|.\] Suppose we now order those conjugacy classes such that \(C_1\) is the conjugacy class containing the identity (which has size 1) and that \(C_2, \ldots, C_t, C_{t+1}, \ldots, C_r\) are the remaining classes, but where \(C_2, \ldots, C_t\) are the trivial classes containing only one element. From what we have said previously we know that \[|Z(G)| = \sum_{i=1}^t |C_i|,\] that is, the size of the centre is equal to the sum of all of the conjugacy classes of size 1 since each element in those classes must be in the centre. This then leads us to the formal definition:

Definition 6.2 (Class Equation) Let \(C_1, C_2, \ldots, C_r\) be the distinct conjugacy classes of a finite group \(G\), ordered so that \(C_1\) is the conjugacy class containing just the identity and \(C_2, \ldots, C_t\) \((t \leq r)\) are any further classes that contain just one element. Then the class equation of \(G\) is \[|G| = |Z(G)| + \sum_{i=t+1}^r |C_i|.\]

In this equation, the summation is the sum of the sizes of all of the non-trivial conjugacy classes. An important property of the Class Equation is that each of the terms on the right hand side is a divisor of the order of \(G\). This follows since \(Z(G)\) is a subgroup of \(G\), so the size of \(Z(G)\) must divide the order of \(G\) and the size of each summand (being an orbit of the action) also divides the order of \(G\).

Example 6.3

The class equation of \(D_3\) is \[\begin{align*} |D_3| &= |\{e\}| + |\{r_1,r_2\}| + |\{s_1,s_2,s_3\}| \\ &= 1 +2 + 3 \\ &= 6. \end{align*}\]

6.2 Finite \(p\)-Groups

Before proceeding with our development of the Sylow Theorems we pause to consider the structure of finite \(p\)-groups (that is, finite groups in which the order of every element is a power of a prime \(p\) (definition) or, equivalently, groups of order \(p^r\) for some prime \(p\) - see Theorem 3.6). We know that groups of order a prime \(p\) are cyclic (and hence abelian). We will soon show that groups of order \(p^2\) are abelian and, therefore, isomorphic to either \(\Z_{p^2}\) or \(\Z_p \times \Z_p\). We first prove that all finite \(p\)-groups have a non-trivial centre (that is, the centre must contain more than one element).

Lemma 6.2 A finite \(p\)-group \(G\) of order \(p^r\) has a non-trivial centre.

Proof.

Consider the class equation for \(G\): \[|G| = |Z(G)| + \sum_{t+1}^{r}|C_i|.\] Now \(p\) divides the order of \(G\) and \(p\) also divides \(\sum_{t+1}^{r}|C_i|\). It follows that \(p\) must divide \(|Z(G)|\). Therefore \(|Z(G)|\) is at least \(p\).

Lemma 6.3 Let \(G\) be a non-abelian group. Then \(G/Z(G)\) is non-cyclic.

Proof.

Suppose \(G/Z(G)\) is cyclic. This means there is an element \(g \in G\) such that \[\{g^i Z(G): i \in \Z\} = G/Z(G)\] that is, the element \(gZ(G)\) generates all of \(G/Z(G)\). This now means that every element of \(G\) can be written in the form \(g^{i}z\) for some \(i \in \Z\) and \(z \in Z(G)\). Now let \(h_1,h_2 \in G\). There are \(i,j \in \Z\) and \(z_1,z_2 \in Z(G)\) such that \(h_1 = g^{i}z_1 =\) and \(h_2 = g^j{z_2}\). We have: \[\begin{align*} h_1h_2 &= g^{i}z_1g^j{z_2} = g^{i}g^{j}z_1z_2 = g^{j}g^{i}z_1z_2\\ &= g^{j}g^{i}z_2z_1 = g^{j}z_2g^{i}z_1 = h_2 h_1. \end{align*}\] Therefore, for all \(h_1,h_2 \in G\) \(h_1 h_2 = h_2 h_1\) and \(G\) is abelian. This yields the desired contradiction.

Theorem 6.1 A group \(G\) of order \(p^2\), where \(p\) is prime, is necessarily abelian.

Proof.

By Lemma 6.2, \(p\) divides the order of \(|Z(G)|\). Therefore the order of \(Z(G)\) is either \(p\) or \(p^2\). If \(|Z(G)| = p^2\), then \(G=Z(G)\) and \(G\) is therefore abelian. We may therefore assume that \(|Z(G)| = p\).

Now observe that if \(|Z(G)| = p\), then \(|G/Z(G)| = p^2/p = p\). Recall that groups of prime order are always cyclic and, hence, abelian. Therefore \(G/Z(G)\) is abelian. Lemma 6.3 now implies that \(G\) must be abelian.

For any prime \(p\), there are exactly two isomorphically distinct groups, \(\Z_{p^2}\) and \(\Z_p \times \Z_p\), of order \(p^2\).

Proof.

The result follows from the Fundamental Theorem of Finite Abelian Groups and the fact that for any prime, \(p\), \(\Z_p \times \Z_p\) is abelian as is \(\Z_{p^2}\) by the above theorem.

Example 6.4

The only groups of order

4 are \(\Z_4\) and \(\Z_2 \times \Z_2\);
9 are \(\Z_9\) and \(\Z_3 \times \Z_3\);
10,201 are \(\Z_{10201}\) and \(\Z_{101}\times \Z_{101}\).

But what about groups of order \(p^3\)?

Let us consider first groups of order \(8 = 2^3\). There are two non-abelian groups of order \(8\): \(D_4\) and \(Q_8\). This means that we need a little more than Theorem 6.1 to classify groups of order \(p^3\).

Notice that every non-abelian group \(G\) of order \(p^3\) must have a centre of order \(p\). This follows since if \(|Z(G)| = p^2\), then \(G/Z(G)\) has order \(p\) and so, by Lemma 6.2, \(G\) must be abelian which would be a contradiction.

Now if \(Z(G)\) has order \(p\), then \(G/Z(G)\) has order \(p^2\) and so is abelian.

Example 6.5

Both non-abelian groups of order \(8\) have centres of order \(2\). The centre of \(D_4\) is \(\{e, r_2\}\), while the centre of \(Q_8\) is \(\{1,-1\}\). (Recall that \(Q_8 = \{\pm 1, \pm i, \pm j, \pm k\} )\).

6.3 Centralisers and Sylow’s First Theorem

In Lemma 6.1 we showed that a group acts on its own elements by conjugation (that is, conjugation is a group action). Whenever we have a group action we have orbits and stabilizers; the orbits of this action are what we have called the conjugacy classes of the group, but what about the stabilizers?

Definition 6.3 (Centraliser) Let \(x\) be an element from a group \(G\). The centraliser of \(x\) in \(G\) is denoted and defined by \[cent_G(x) = \{ g \in G \, | \, gxg^{-1} = x \} = \{ g \in G \, | \, gx = xg \}.\]

So, the centraliser of \(x\) in \(G\) consists of all of the elements of \(G\) that commute with \(x\). Notice that \(cent_G(x) = G\) if and only if \(x\) is in the centre of \(G\)

Note

Do NOT confuse the centre of a group with the centraliser of an element.

If \(G\) is abelian, \(\centg(x) = G\) for all \(x \in G\). However, for non-abelian group, this only happens when \(x\) is in the centre of \(G\). For example, in \(D_3\): \[\begin{align*} \centg(e) &= D_3 \\ \centg(r_1) &= \{e,r_1,r_2\}=\centg(r_2) \\ \centg(s_1) &= \{e, s_1\} \\ \centg(s_2) &= \{e,s_2\} \\ \centg(s_3) &= \{e, s_3\} \end{align*}\]

Lemma 6.4 Let \(x\) be an element from a group \(G\). Then \(cent_G(x)\) is a subgroup of \(G\).

Proof.

We give a direct and indirect proof.

Indirect proof: The \(\centg(x)\) is the stabiliser of \(x\) under the action of \(G\) on itself by conjugation. By Theorem 3.2, \(\centg(x)\) is therefore a subgroup of \(G\).
Direct proof: we carry out the subgroup checks.
- Clearly \(e \in \centg(x)\) since \(ex = xe\).
- Let \(g,h \in \centg(x)\). Then \[ghx = gxh = xgh\] and so \(gh \in \centg(x)\).
- Let \(g \in \centg(x)\). Then \(gx = xg\). Post and pre-multiplying by \(\ginv\) we have: \[x\ginv = \ginv g x \ginv = \ginv x g \ginv = \ginv x\] and so \(\ginv \in \centg(x)\). This complete the subgroup checks and so \(\centg(x)\) is a group.

Example 6.6

The conjugacy classes in \(D_3\) are \(\{e\}\), \(\{r_1,r_2\}\) and \(\{s_1,s_2,s_3\}\). Notice the following: \[\begin{align*} |\centg(e)|\times |\conjg(e)| &= |D_3| \times |\{e\}| = 6 \\ |\centg(r_1)|\times |\conjg(r_1)| &= |\{e,r_1,r_2\}| \times |\{r_1,r_2\}| = 6 \\ |\centg(s_i)|\times |\conjg(s_i)| &= |\{e,s_i\}| \times |\{s_1,s_2,s_3\}| = 6. \end{align*}\] This is not a coincidence, merely an application of the Orbit-Stabiliser Theorem (Theorem 3.3) applied to the conjugation action of: \[ |\stab(x)| \times |\orb(x)| = |G|.\] Under the conjugacy action, \(\stab(x) = \centg(x)\) and \(\conjg(x) = \orb(x)\).

We are now in a position to state and prove Sylow’s First Theorem.

Theorem 6.2 (Sylow’s First Theorem) Let G be a finite group of order \(p^{\alpha}s\), where \(p\) is a prime not dividing \(s\). Then \(G\) has a subgroup of order \(p^{\beta}\) for each integer \(\beta\) such that \(0 \leq \beta \leq \alpha\).

Proof.

We prove this result by induction on the order of \(G\).

Our base case occurs when \(|G| = 1\). In this case the result is readily seen to hold.

Next assume that that \(|G| = p^{\alpha}s\) (for a prime \(p\) and \(\gcd(s,p) = 1\)) and the result holds for all groups of order strictly less than \(p^{\alpha}s\). The aim is to get to a point in which we can apply the inductive step. Our key tool is the Class equation Definition 6.2: \[|G| = |Z(G)| + \sum_{i=t+1}^r |C_i|.\]

There are two cases to consider: either \(p\) does not divide \(|C_i|\) for some \(i\) or for any \(i\) between \(t+1\) and \(r\), \(p\) divides \(|C_i|\). We consider the first case first.

Case 1: : Let \(i\) be such that \(p\) does not divide \(|C_i|\) and let \(x \in C_i\). By the Orbit-Stabiliser Theorem \(|C_i| \times |\centg(x)| = |G|\). Therefore \(p^{\alpha}\) divides \(|\centg(x)|\) since \(p^{\alpha}\) does not divide \(|C_i|\). Notice moreover that since \(x \notin \Z(G)\), \(\centg(x) \ne G\) and so \(|\centg(x)|< |G|\). Therefore we may apply the inductive hypothesis to conclude the the result holds for \(\centg(x)\). Since \(\centg(x) < G\), then the result holds in \(G\) as well.
Case 2: In this case \(p\) divides the order of \(\Z(G)\), since \(p\) divides \(|C_i\) for all \(i\) and \(p\) divides \(|G|\). We may now apply Cauchy’s Theorem ( Theorem 3.5) to conclude that \(\Z(G)\) has a subgroup \(N\) of order \(p\). Notice that as \(N \le Z(G)\), then \(N\) is a normal subgroup of \(G\) (\(g n \ginv = n\) for all \(g \in G\) and \(n \in N\)). Now as \(G/N\) has order \(p^{\alpha -1}s\) we may therefore apply the induction hypothesis to \(G/N\). Let \(\beta\) be a number between \(0\) and \(\alpha -2\). By induction \(G/N\) has a subgroup of order \(p^{\beta}\). Applying the correspondence theorem (Theorem 4.5), there is therefore a subgroup \(H \le G\) such that \(H/N\) is a subgroup of \(G/N\) of order \(p^{\beta}\). Now observe that \(H\) must have have order \(p^{\beta +1}\) since, if \(H/N = \{N, h_2N, \ldots, h_{p^{\beta}}N \}\), then \[H = N \sqcup h_2N \sqcup \ldots \sqcup h_{p^{\beta}}N\] has order \(p^{\beta +1}\). Therefore, for all \(i\) between \(0\) and \(\alpha\), \(G\) has a subgroup of order \(p^{i}\) (the trivial subgroup \(\{e\}\) of \(G\) is a subgroup of order \(1\).)

Definition 6.4 (Sylow \(p\)-subgroup) Let \(G\) be a finite group of order \(p^{\alpha}s\), where \(p\) is a prime not dividing \(s\). A subgroup of \(G\) of order \(p^{\alpha}\) is called a Sylow p-subgroup of \(G\).

Example 6.7

We consider some groups we have already encountered:

\(|D_3| = 2 \times 3\). The Sylow-2 subgroups are \(\{e, s_i\}\), \(i=1,2,3\). The only Sylow-3 subgroup is \(\{e, r_1, r_2\}\).
\(|A_4| = 12 = 4 \times 3\). Thus \(A_4\) has Sylow 3-subgroups of order \(3\) (each generated by a three-cycle) and a Sylow 2-subgroup of order \(4\) namely \(\{\id, (1 \; 2) (3 \; 4),(1 \; 3)(2 \; 4),(1 \; 4)(2 \;3) \}\).

The notion of conjugacy can be extended to subsets of a group \(G\), though in practice only subgroups are of importance.

Definition 6.5 Let \(H\) be a subgroup of a group \(G\), and \(g \in G\). The subset \(gHg^{-1} = \{ ghg^{-1} \, | \, h \in H \}\) is called the conjugate of \(H\) by \(g\).

We have seen this idea before when we proved Theorem 4.6 using group actions, and needed the fact that if \(H\) is a subgroup then \(gHg^{-1}\) is a subgroup. The following Lemma tells us even more.

Lemma 6.5 Let \(H\) be a subgroup of a group \(G\) and \(g \in G\). Then \(gHg^{-1}\) is a subgroup of \(G\) isomorphic to \(H\).

Proof.

First we show that \(gH\ginv\) is a subgroup.

Clearly, \(e \in g H \ginv\) since \(g e \ginv = e\).

Let \(h_1,h_2 \in H\). Then

\[gh_1\ginv g h_2 \ginv = gh_1 h_2 \ginv \in gH\ginv.\] Therefore \(gH\ginv\) is closed under products.

Lastly let \(h \in H\), then \[(gh\ginv)^{-1} = g h^{-1}\ginv \in gH \ginv\] and so \(gh\ginv\) is closed under inverses.

The map \(\phi: H \to gH\ginv\) by \(h \mapsto gh\ginv\) is an isomorphism.

Clearly \(\phi\) is surjective. To see that is is injective observe that if \[gh_1\ginv = gh_2 \ginv\] then, post-multiplying by \(g\) and pre-multiplying by \(\ginv\) we have: \[h_1 = \ginv g h_1 \ginv g = \ginv g h_2 \ginv g = h_2.\] Therefore \(\phi\) is a bijection.

To verify that \(\phi\) is a homomorphism, observe \[\phi(h_1h_2) = gh_1h_2 \ginv = gh_1 \ginv gh_2 \ginv = \phi(h_1)\phi(h_2)\] as required.

There is an immediate connection between conjugacy and normal subgroups; \(H\) is a normal subgroup of \(G\) if and only if for all \(g \in G, \; gH = Hg\), but \(gH = Hg \ifif gHg^{-1} = H\). It follows that \(H\) is normal in \(G\) if and only if \(H\) has no distinct conjugates in \(G\), that is we only get \(H\). Note that if \(H\) is any subgroup of \(G\) then if \(H\) is normal in \(G\) we have that \(gHg^{-1}=H \; \forall g \in G\); if \(H\) is not normal in \(G\) then \(\exists \, x \in G\) such that \(xHx^{-1} \neq H\) but \(xHx^{-1}\) is still a subgroup of \(G\).

We can now state the remaining Sylow Theorems.

Theorem 6.3 (Sylow’s Second Theorem) Let \(H_1\) and \(H_2\) be Sylow \(p\)-subgroups of a finite group \(G\). Then \(H_1\) and \(H_2\) are conjugate.

That is, there exists \(x \in G\) such that \(xH_1x^{-1}=H_2\) and vice versa.

Theorem 6.4 (Sylow’s Third Theorem) The number of Sylow \(p\)-subgroups of a finite group \(G\) is congruent to 1 modulo \(p\) and divides \(|G|\).

We now consider an application.

Let \(G\) be a group of order \(3 \times 5 = 15\). By Cauchy’s Theorem \(G\) has at least one subgroup of order \(3\) and at least one of order \(5\).

Let \(n_3\) be the number of subgroups of order \(3\). By Sylow’s Third Theorem \(n_3 \equiv 1 \pmod{3}\) and divides the order of \(G\) which is \(15\). Therefore \(n_3\) must be \(1\) since \(5 \equiv 2 \pmod 3\). Therefore \(G\) has a unique Sylow \(3\)-subgroup, \(N_1\). By Sylow’s Second Theorem, \(N_1\) must be normal in \(G\) since it is conjugate only to itself.

In a similar way \(G\) has a normal subgroup \(N_2\) of order \(5\). Notice that \(N_1 \cap N_2 = \{e\}\) since every non-trivial element of \(N_1\) has order \(3\) and every non-trivial element of order \(N_2\) has order \(5\).

Let \(a\) be a non-trivial element of \(N_1\) and \(b\) be a non-trivial element of \(N_2\) and consider the product \(ab\). Now \(ab\) has order \(1,3, 5\) or \(15\).

If the order of \(ab =1\), then \(ab = e\) and so \(b = a^{-1}\) which contradicts the fact that \(N_1 \cap N_2 = \{e\}\).

If the order of \(ab = 3\), then \(\gen{ab}\) must be \(N_1\) (since \(H_1\) is the only Sylow \(3\)-subgroup of \(G\)). This means that \(ab = n \in N_1\) and so \(b = a^{-1}n \in N_1\) which contradicts the fact that \(N_1 \cap N_2 = \{e\}\).

If the order of \(ab = 5\), then as in the previous case \(ab \in N_2\) and \(a = mb^{-1}\) for some \(m \in N_2\) and so \(a \in N_2\) yielding a contradiction.

We must therefore conclude that \(ab\) has order \(15\) and so \(G\) is a cyclic group.

Note that the above does not generalise to groups of order \(pq\), for distinct primes \(p\) and \(q\) since there are non-abelian groups of order \(pq\) (e.g . \(D_p\) for every prime \(p\) is a non-abelian group of order \(2p\)). In the case where a group \(G\) of order \(p,q\) has a unique Sylow \(p\)-subgroup and a unique Sylow \(q\)-subgroup, the argument above does generalise.

Let us suppose that \(p\) and \(q\) are distinct primes with \(q<p\) and consider a group \(G\) or order \(pq\).

There is always a unique Sylow \(p\) subgroup, since \(n_p\), the number of Sylow \(p\) subgroups must divide \(pq\) and be congruent to \(1\) modulo \(p\) i.e. \(n_p\) divides \(q\) and is congruent to \(1\) modulo \(p\). As an example, consider \(D_3\). This has order \(2\times 3\) and a unique (and so normal) Sylow \(3\)-subgroup \(\{e,r_1,r_2\}\).

What about \(n_q\) the number of Sylow \(q\)-subgroups of \(G\)? We know that \(n_q \equiv 1 \pmod q\) and \(n_q\) divides \(pq\). This means that \(n_q\) is either \(1\) or \(p\). If \(p\) is not congruent to \(1\) modulo \(q\), then \(n_q = 1\) and \(G\) is cyclic.

If on the other hand \(p \equiv 1 \pmod q\), then \(n_q\) might be equal to \(p\). For example if we consider the \(D_p\), here \(q=2\), then \(D_p\) has \(p\) Sylow \(2\)-subgroups: \(\{e, s_i\}\), \(i=1,2,\ldots, p\).

We can summarise the above as follows:

Let \(p\) and \(q\) be distinct primes with \(q<p\) and \(p \not \equiv 1 \pmod q\). Then a group \(G\) of order \(pq\) is cyclic.

Applying this to groups of order \(n = pq\) for small \(n\) we have:

\[ \begin{array}{c|c|c|c} n & q & p & \\ \hline 15 & 3 & 5& \text{cyclic only }\\ 21 & 3 & 7& \text{possibly non-cyclic }\\ 33 & 3 & 11 & \text{cyclic only } \\ 35 & 5 & 7& \text{cyclic only }\\ 55 & 5 & 11& \text{possibly non-cyclic}\\ 77 & 7 & 11& \text{cyclic only }\\ 91 & 7 & 13& \text{cyclic only }\\ 119 & 7 & 17& \text{cyclic only } \end{array} \]

6.4 Problem Sheet 6

Covers Chapter 6.

Question 6.1

Determine whether the converse of Lagrange’s Theorem is true for all finite \(p\)-groups. Provide either a proof or a counter-example to justify your assertion.

Question 6.2

Show that every group of order 63 has a normal subgroup of order 7 and at least one subgroup of order 21.

Question 6.3

Use the Sylow Theorems to show that every group of order 980 has a normal subgroup of order 49, and subgroups of orders 245 and 98.

Question 6.4

Let \(G\) be a group of order \(pq\), where \(p\) and \(q\) are primes, \(p<q\), and \(p\) does not divide \(q-1\).

List the possible orders of such groups up to and including 40.
Prove that \(G\) contains normal subgroups of orders \(p\) and \(q\).
Let \(H = \langle x \rangle\) and \(K = \langle y \rangle\) be normal subgroups of orders \(p\) and \(q\) respectively. Show that \(x(y \xinv \yinv) \in H\) and \((xy\xinv)\yinv\in K\). [Hint: the brackets are supposed to be helpful…]
Deduce that \(xy = yx\). [Hint: what is \(H \cap K\)?]
Determine the order of the element \(xy\) in \(G\) and deduce that \(G\) is cyclic.

Solution 6.4

\(3 \times 5 = 15\), \(3 \times 11 = 33\), \(5 \times 7 = 35\).
Let \(n_p\) be the number of Sylow \(p\)-subgroups of order \(p\) and \(n_q\) be the number of Sylow \(q\)-subgroups of order \(q\). By Sylow’s Third Theorem, \(n_p = 1 + kp\) for some integer \(k\), \(n_q = 1 + mq\) for some integer \(m\) and both \(n_p\) and \(n_q\) divide \(pq\). It then follows that both \(n_p\) and \(n_q\) are equal to 1, \(p\) or \(q\) (we dismiss the case \(pq\) since that is impossible as there is only one normal subgroup with the same order as the group, namely the group itself). Now, \(n_p = 1 + kp \neq p\) and \(n_q = 1 + mq \neq q\). Also, \(p<q\) so \(n_q \neq p\). If \(n_p = q\) then \(1 + kp = q \imp q-1\) is a multiple of \(p\) and this is not allowed. Hence we have that \(n_p = n_q = 1\) and, therefore, both Sylow subgroups are normal.
Note that \(x \in H\) and so \(\xinv \in H\). Also, \(H\) is normal in \(G\) and so for all \(y \in G\) we have that \(y\xinv \yinv \in H\). Hence, since \(H\) is closed, \(x(y \xinv \yinv) \in H\). Similarly, \(y, \yinv \in K\) and K is normal in \(G\), so for all \(x \in G\) we have \((xy\xinv) \in K\) and, hence \((xy\xinv)\yinv \in K\).
Since all non-identity elements of \(H\) have order \(p\) and all non-identity elements of \(K\) have order \(q\) then it follows that \(H \cap K = \{e\}\). But in (c) above we showed that \(xy\xinv \yinv \in H\) and \(xy\xinv \yinv \in K\), so \(xy\xinv \yinv \in H \cap K\). Hence \(xy\xinv \yinv = e\) and so \[xy = (\xinv \yinv)^{-1} = (\yinv)^{-1}(\xinv)^{-1} = yx.\]
We know that \(xy\) has order \(1, p, q\) or \(pq\). We also know that \(H \cap K = \{e\}\). If \(xy\) has order 1 then \(x = \yinv \in K\). This is a contradiction as \(x \notin K\) since it has order \(p\). If \(xy\) has order \(p\) then \(xy \in H\) and so \(xy = x^i\) for some \(i \in \Z\). Then \(y = x^{i-1} \in H\) which is a contradiction since \(y \notin H\). If \(xy\) has order \(q\) then, similarly, \(x \in K\) which is a contradiction. It then follows that \(xy\) has order \(pq\) and, since that is the order of \(G\), then \(G\) must be cyclic.